You just use the * operator, as you might expect, e.g.

y = x * k;

where x is the input image, k is a scalar value, and y is the resulting scaled image.

v*u is algebraic matrix multiplication, for which the rule is that for an (M x N) * (P x Q) operation, that N must equal P and the output size is M x Q -- so (M x N) * (N x Q) giving M x Q. You have (1 x 5) * (1 x 5) which violates that rule. When you take u * v' then you have (1 x 5) * (5 x 1) giving 1 x 1. v.*u would go element by element, result(K) = v(K) * u(K) which would be 1 x 5 result.

You need to vectorise the division (replace *|‘/’|* with *|‘./’|*). Then it works: ... CODE .. %Define Dependent Variables n_I = (3/2)*(d_p / d_c).^2; n_G = ((rho_p - rho_w)*g*(d_p).^2)./(18 * mu * v_f); n_D = 0.9*((k_B*T)./(mu * d_p * d_c * v_f)).^(2/3); X = n_I + n_G + n_D; %Define C/Co C = exp(((-3/2)*(1-E)*X*a*L)/d_c); %Plot C/Co plot(v_f,C); xlabel('Filtration Velocity (m/h)'); ylabel('C/Co'); title(sprintf('Filter Performance with %.1f \\mum', d_f)) I didn’t copy all your code to my Answer, only the part that needed help. I also got the impression you wanted *|‘d_f’|* to be displayed in the *|title|*, so I added that.

bsxfun(@times,[a b],[x1 y1; x2, y2; x3 y3]) here |a, b, x1, y1, x2, y2,x3, y3| - numeric